我对Java比较陌生,经常发现需要对值进行Map<Key,Value>排序。
由于这些值不是唯一的,我发现自己将keySet转换为一个数组,并使用自定义比较器通过数组排序对该数组进行排序,该比较器根据与该键关联的值进行排序。
有没有更简单的方法?
我对Java比较陌生,经常发现需要对值进行Map<Key,Value>排序。
由于这些值不是唯一的,我发现自己将keySet转换为一个数组,并使用自定义比较器通过数组排序对该数组进行排序,该比较器根据与该键关联的值进行排序。
有没有更简单的方法?
当前回答
迟到。
随着Java-8的出现,我们可以以非常简单/简洁的方式使用流进行数据操作。您可以使用流按值对映射条目进行排序,并创建一个LinkedHashMap,以保留插入顺序迭代。
Eg:
LinkedHashMap sortedByValueMap = map.entrySet().stream()
.sorted(comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)) //first sorting by Value, then sorting by Key(entries with same value)
.collect(LinkedHashMap::new,(map,entry) -> map.put(entry.getKey(),entry.getValue()),LinkedHashMap::putAll);
对于反向排序,请替换:
comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)
with
comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey).reversed()
其他回答
我合并了user157196和Carter Page的解决方案:
class MapUtil {
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue( Map<K, V> map ){
ValueComparator<K,V> bvc = new ValueComparator<K,V>(map);
TreeMap<K,V> sorted_map = new TreeMap<K,V>(bvc);
sorted_map.putAll(map);
return sorted_map;
}
}
class ValueComparator<K, V extends Comparable<? super V>> implements Comparator<K> {
Map<K, V> base;
public ValueComparator(Map<K, V> base) {
this.base = base;
}
public int compare(K a, K b) {
int result = (base.get(a).compareTo(base.get(b)));
if (result == 0) result=1;
// returning 0 would merge keys
return result;
}
}
我建议使用Arrays.sort,而不是像某些人那样使用Collections.ort。实际上Collections.ort的作用是这样的:
public static <T extends Comparable<? super T>> void sort(List<T> list) {
Object[] a = list.toArray();
Arrays.sort(a);
ListIterator<T> i = list.listIterator();
for (int j=0; j<a.length; j++) {
i.next();
i.set((T)a[j]);
}
}
它只调用列表上的array,然后使用Arrays.sort。这样,所有映射条目将被复制三次:一次从映射复制到临时列表(无论是LinkedList还是ArrayList),然后复制到临时数组,最后复制到新映射。
我的解决方案省略了这一步,因为它不会创建不必要的LinkedList。以下是代码,通用友好,性能最佳:
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map)
{
@SuppressWarnings("unchecked")
Map.Entry<K,V>[] array = map.entrySet().toArray(new Map.Entry[map.size()]);
Arrays.sort(array, new Comparator<Map.Entry<K, V>>()
{
public int compare(Map.Entry<K, V> e1, Map.Entry<K, V> e2)
{
return e1.getValue().compareTo(e2.getValue());
}
});
Map<K, V> result = new LinkedHashMap<K, V>();
for (Map.Entry<K, V> entry : array)
result.put(entry.getKey(), entry.getValue());
return result;
}
好的,这个版本使用两个新的Map对象和两次迭代,并对值进行排序。希望,虽然地图条目必须循环两次,但表现良好:
public static void main(String[] args) {
Map<String, String> unsorted = new HashMap<String, String>();
unsorted.put("Cde", "Cde_Value");
unsorted.put("Abc", "Abc_Value");
unsorted.put("Bcd", "Bcd_Value");
Comparator<String> comparer = new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
return o1.compareTo(o2);
}};
System.out.println(sortByValue(unsorted, comparer));
}
public static <K, V> Map<K,V> sortByValue(Map<K, V> in, Comparator<? super V> compare) {
Map<V, K> swapped = new TreeMap<V, K>(compare);
for(Entry<K,V> entry: in.entrySet()) {
if (entry.getValue() != null) {
swapped.put(entry.getValue(), entry.getKey());
}
}
LinkedHashMap<K, V> result = new LinkedHashMap<K, V>();
for(Entry<V,K> entry: swapped.entrySet()) {
if (entry.getValue() != null) {
result.put(entry.getValue(), entry.getKey());
}
}
return result;
}
该解决方案使用带有比较器的TreeMap,并对所有空键和值进行排序。首先,使用TreeMap中的排序功能对值进行排序,然后使用排序后的Map创建一个结果,因为LinkedHashMap保留了相同的值顺序。
格里兹,GHad
基于@devinmore代码,一种使用泛型并支持升序和降序排序的map排序方法。
/**
* Sort a map by it's keys in ascending order.
*
* @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
* @author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map) {
return sortMapByKey(map, SortingOrder.ASCENDING);
}
/**
* Sort a map by it's values in ascending order.
*
* @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
* @author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map) {
return sortMapByValue(map, SortingOrder.ASCENDING);
}
/**
* Sort a map by it's keys.
*
* @param sortingOrder {@link SortingOrder} enum specifying requested sorting order.
* @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
* @author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map, final SortingOrder sortingOrder) {
Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
public int compare(Entry<K, V> o1, Entry<K, V> o2) {
return comparableCompare(o1.getKey(), o2.getKey(), sortingOrder);
}
};
return sortMap(map, comparator);
}
/**
* Sort a map by it's values.
*
* @param sortingOrder {@link SortingOrder} enum specifying requested sorting order.
* @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
* @author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map, final SortingOrder sortingOrder) {
Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
public int compare(Entry<K, V> o1, Entry<K, V> o2) {
return comparableCompare(o1.getValue(), o2.getValue(), sortingOrder);
}
};
return sortMap(map, comparator);
}
@SuppressWarnings("unchecked")
private static <T> int comparableCompare(T o1, T o2, SortingOrder sortingOrder) {
int compare = ((Comparable<T>)o1).compareTo(o2);
switch (sortingOrder) {
case ASCENDING:
return compare;
case DESCENDING:
return (-1) * compare;
}
return 0;
}
/**
* Sort a map by supplied comparator logic.
*
* @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
* @author Maxim Veksler
*/
public static <K, V> LinkedHashMap<K, V> sortMap(final Map<K, V> map, final Comparator<Map.Entry<K, V>> comparator) {
// Convert the map into a list of key,value pairs.
List<Map.Entry<K, V>> mapEntries = new LinkedList<Map.Entry<K, V>>(map.entrySet());
// Sort the converted list according to supplied comparator.
Collections.sort(mapEntries, comparator);
// Build a new ordered map, containing the same entries as the old map.
LinkedHashMap<K, V> result = new LinkedHashMap<K, V>(map.size() + (map.size() / 20));
for(Map.Entry<K, V> entry : mapEntries) {
// We iterate on the mapEntries list which is sorted by the comparator putting new entries into
// the targeted result which is a sorted map.
result.put(entry.getKey(), entry.getValue());
}
return result;
}
/**
* Sorting order enum, specifying request result sort behavior.
* @author Maxim Veksler
*
*/
public static enum SortingOrder {
/**
* Resulting sort will be from smaller to biggest.
*/
ASCENDING,
/**
* Resulting sort will be from biggest to smallest.
*/
DESCENDING
}
public class Test {
public static void main(String[] args) {
TreeMap<Integer, String> hm=new TreeMap();
hm.put(3, "arun singh");
hm.put(5, "vinay singh");
hm.put(1, "bandagi singh");
hm.put(6, "vikram singh");
hm.put(2, "panipat singh");
hm.put(28, "jakarta singh");
ArrayList<String> al=new ArrayList(hm.values());
Collections.sort(al, new myComparator());
System.out.println("//sort by values \n");
for(String obj: al){
for(Map.Entry<Integer, String> map2:hm.entrySet()){
if(map2.getValue().equals(obj)){
System.out.println(map2.getKey()+" "+map2.getValue());
}
}
}
}
}
class myComparator implements Comparator{
@Override
public int compare(Object o1, Object o2) {
String o3=(String) o1;
String o4 =(String) o2;
return o3.compareTo(o4);
}
}
输出,输出=
//sort by values
3 arun singh
1 bandagi singh
28 jakarta singh
2 panipat singh
6 vikram singh
5 vinay singh